On 7 April in DR position LAT 27°42.0'N, LONG 114°03.0'W, you observe an amplitude of the Sun. The Sun's center is on the celestial horizon and bears 076°psc. The chronometer reads 02h 10m 17s and is 01m 52s slow. Variation in the area is 8°E. What is the deviation of the standard magnetic compass?

Question

On 7 April in DR position LAT 27&deg;42.0'N, LONG 114&deg;03.0'W, you observe an amplitude of the Sun. The Sun's center is on the celestial horizon and bears 076&deg;psc. The chronometer reads 02h 10m 17s and is 01m 52s slow. Variation in the area is 8&deg;E. What is the deviation of the standard magnetic compass?

Accepted Answer

Option A: 1.8&deg;W. Determine the true amplitude using the formula sin(Amp) = sin(Dec) / cos(Lat). Calculate the Sun's declination for the date, then find the true bearing. Comparing the true bearing to the compass bearing (076 degrees psc) gives the compass error. Adjusting for 8 degrees East variation results in 1.8 degrees West deviation.